﻿using System.Collections.Generic;
using System.Drawing;

namespace ProblemsSet
{
    public class Problem_65 : BaseProblem
    {
        public override object GetResult()
        {
            var index = 100;

            var cfr = new List<long>(index);

            for (int i = 1; i <= index; i++)
            {
                cfr.Add(GetKoefForE(i));
            }

            var nom = "";
            var denom = "";
            long qqq = 0;
            MathLogic.GetDivideByFractions(cfr, out nom, out denom);
            foreach (var c in nom)
            {
                qqq += c - 48;
            }

            return qqq;
        }

        private static long GetKoefForE(long index)
        {
            if (index == 1) return 2;
            if (index % 3 != 0) return 1;
            return 2*(index/3);
        }

        public override string Problem
        {
            get
            {
                return @"The square root of 2 can be written as an infinite continued fraction.

2 = 1 +	
1

 	2 +	
1

 	 	2 +	
1

 	 	 	2 +	
1

 	 	 	 	2 + ...
The infinite continued fraction can be written, 2 = [1;(2)], (2) indicates that 2 repeats ad infinitum. In a similar way, 23 = [4;(1,3,1,8)].

It turns out that the sequence of partial values of continued fractions for square roots provide the best rational approximations. Let us consider the convergents for 2.

1 +	
1

= 3/2
 	
2
 
1 +	
1

= 7/5
 	2 +	
1

 	 	
2
 
1 +	
1

= 17/12
 	2 +	
1

 
 	 	2 +	
1

 
 	 	 	
2
 
1 +	
1

= 41/29
 	2 +	
1

 	 	2 +	
1

 
 	 	 	2 +	
1

 
 	 	 	 	
2
 
Hence the sequence of the first ten convergents for 2 are:

1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, 3363/2378, ...
What is most surprising is that the important mathematical constant,
e = [2; 1,2,1, 1,4,1, 1,6,1 , ... , 1,2k,1, ...].

The first ten terms in the sequence of convergents for e are:

2, 3, 8/3, 11/4, 19/7, 87/32, 106/39, 193/71, 1264/465, 1457/536, ...
The sum of digits in the numerator of the 10th convergent is 1+4+5+7=17.

Find the sum of digits in the numerator of the 100th convergent of the continued fraction for e.";
            }
        }

        public override bool IsSolved
        {
            get
            {
                return true;
            }
        }

        public override object Answer
        {
            get
            {
                return 272;
            }
        }

    }
}
